Ways to prove $ \int_0^\pi \frac{\sin^2 nx}{\sin^2 x} dx= n\pi$ In how many ways can we prove the following theorem?
$$I(n):= \int_0^\pi \frac{\sin^2 nx}{\sin^2 x} dx= n\pi$$
where $n$ is a nonnegative integer.
The proof I found is by considering $I(n+1)-I(n)$, which can be reduced to
$$
g(n):= \int_0^\pi \frac{\sin(2 n x) \cos x}{\sin x}dx
$$
I then showed that $g(n)=g(n+1)$, with $g(n) = ng(1) = n\pi$. This completes the proof.
I was wondering if there is a more direct way to prove it. By 'direct' I mean without deriving auxiliary recursions.
 A: Here is a cannon to shoot a fly.
Rewrite the integral as
\begin{equation}
I(n):= \int_0^\pi\frac{1-\cos2nx}{1-\cos2x}\,dx\stackrel{2x\,\mapsto\, x}\Longrightarrow \frac{1}{2}\int_0^{2\pi}\frac{1-\cos nx}{1-\cos x}\,dx\tag{1}
\end{equation}
From my answer here, we have
\begin{equation}\int_0^{2\pi}\frac{\cos mx}{p-q\cos x}\, dx=\frac{2\pi}{\sqrt{p^2-q^2}}\left(\frac{p-\sqrt{p^2-q^2}}{q}\right)^m\qquad\hbox{for}\qquad |q|<p\tag{2}
\end{equation}
Now, we will treat the integral $(1)$ as though it is separated using $(2)$. We must be careful here because each integrals diverge. We set $m=0$, $m=n$, $p=1$, and take the limit as $q\to1^-$, then
\begin{align}
I(n)&:= \lim_{q\to1^-}\left[\;\frac{\pi}{\sqrt{1-q^2}}-\frac{\pi}{\sqrt{1-q^2}}\left(\frac{1-\sqrt{1-q^2}}{q}\right)^n\;\right]\tag{3}
\end{align}
The limit above succumbs to apply L'Hôpital's once, then it follows
\begin{equation}
I(n):= \int_0^\pi\frac{\sin^2nx}{\sin^2x}\,dx=n\pi
\end{equation}
which is the announced result.$\qquad\square$
A: I can post my solution , from my document. It can be found on page 83. If one truly want to shoot this problem with a canon one can use the following generalization
$$
\int _{0}^{\pi }\! \left( {\frac {\sin \left( nx \right) }{\sin
 \left( x \right) }} \right)^{m}{dx}=\pi \sum _{l=0}^{\large\left\lfloor {\frac{m\left(n-1\right)}{2n}} \right\rfloor }\left( -1 \right) ^{l}{m\choose l}{\dfrac{m}{2}\left( n+1\right) -ln-1\choose m-1}\tag{10}
$$
Which is proved by Graham Hesketh, see equation 10  here. Set $m=1$ use the properties of the floor function and one is is done.
Another way to prove $(n)$ can be found in the same answer

Proof

Lemma: Let $k \in \mathbb{Z}$ then
$$\begin{align*}
\int_0^{\pi}\frac{\sin 2kx}{\sin x}\mathrm{d}x & = 0 \tag{1} \\ \int_0^{\pi}\frac{\sin (2k-1)x}{\sin x}\mathrm{d}x & = \pi \tag{2} \end{align*}$$

Proof:  We first define the following function $\displaystyle I_n = \int_0^\pi \frac{\sin 2n x}{\sin x} \mathrm{d}x$. Note that we now have $I_0 = 0$ and $I_1=\pi$. since $\sin 0 = 0$ and $\sin 2x = 2\cos x \sin x$. We have the following relation for all $n$
$$
I_n-I_{n-2}
=\int_0^{\pi}\frac{\sin{nx}-\sin{(n-2)x}}{\sin x}\mathrm{d}x
=2\int_0^{\pi}\cos(n-1)x\mathrm{d}x
=2\left[\frac{\sin{(n-1)x}}{n-1}\right]_0^{\pi}
=0
$$
For $|n|\geq 3$. This means $I_{2k}=I_{2k-2}=\cdots=I_{2}=\pi$ and simmilarly $I_{2k+1} = I_{2k-1}=\cdots I_1=0$. Which is what we wanted to show.

Proposition: Let $k \in \mathbb{Z}$ then
$$
\ell_2(k) 
= \int_0^\pi \left( \frac{\sin kx}{\sin x} \right)^2\mathrm{d}x
= \int_0^\pi \frac{1 - \cos kx}{1 - \cos x}\,\mathrm{d}x
= |k|\pi \tag{3}
$$

Proof 1: Here we will follow in the footsteps of the other answer and use the lemma above. We define $\displaystyle J_n=\int_0^{\pi}\left(\frac{\sin nx}{\sin x}\right)^2\mathrm{d}x$. We can note that $J_n - J_{n-1}$ is constant eg
$$ 
\begin{align*}
    J_k - J_{k-1}
& = -\frac{1}{2}\int_0^{\pi}\frac{\cos{2kx}-\cos{(2n-2)x}}{(\sin x)^2}
  = -\frac{1}{2}\int_0^{\pi}\frac{-2\sin\left(\frac{4k-2}{2}x\right) 
    \sin\left(\frac{2x}{2}\right)}{(\sin x)^2} \\
& = \int_0^{\pi}\frac{\sin{(2k-1)x}}{\sin x}
  = I_{2k-1}=\pi.
\end{align*}
$$
Then we can write
$$J_n=J_{n-1}+\pi=J_{n-2}+2\pi=\cdots=J_1+(n-1)\pi=n\pi.$$
This completes the proof.

Proof 2: Here is another way to solve this problem. I got the idea from chat, but the tecnique is much older. We want to prove that $(I_{n+1} + I_{n-2}) / 2 = I_n$. Eg that $I_n$ is the average of the next and previous term
Some calculations show that
\begin{align*}
     \frac{I_{n+1}+I_{n-1}}{2}
 & = \frac{1}{2}\int_0^\pi \frac{1 - \cos(n+1)x }{1 - \cos x} + \frac{1-\cos(n-1)x}{1-\cos x} \,\mathrm{d}x \\
 & = \frac{1}{2}\int_0^\pi \frac{2 -\bigl[ \cos(n+1)x + \cos(n-1)x\bigr]}{1-\cos x} \,\mathrm{d}x \\
 & =  \int_0^\pi \frac{1 - \cos nx \cos x}{1-\cos x} \,\mathrm{d}x \\
 & =  \int_0^\pi \frac{(1-\cos nx) +  (1-\cos x) \cos nx}{1-\cos x} \,\mathrm{d}x \\
 & =  \int_0^\pi \frac{1 - \cos nx}{1-\cos x} \,\mathrm{d}x  
   = I_n
\end{align*}
Several things was used here. Like $\cos (n+1)x + \cos(n-1)x = 2 \cos nx \cos x$ and $ \int_0^\pi \cos n x\,\mathrm{d}x = 0 \ \forall \ n \in\mathbb{Z} \backslash \{0\}$. Now we have shown that
$$
 I_n = \frac{I_{n+1}+I_{n-1}}{2}
 \ \Rightarrow \
 2I_n = I_{n+1}+I_{n-1}
 \ \Rightarrow \
 I_n - I_{n-1} = I_{n+1} - I_n
$$
Which is just an arithmetric sequence because the difference between two terms is constant (same as the previous answer). Hence
$I_n = I_0 + (n-0)d = nd$, where $d$ is the difference between two terms. We have $d = I_1 - I_0 = \pi - 0 = \pi$. This completes the second proof. $\qquad\square$
A: Recall that the Féjer kernel $F_n(t)$ is defined to be the mean of the Dirichlet kernel $$D_n(t)=\sum_{|k|\leqslant n}e^{ikt}$$ 
It is known $F_n(2t)=\dfrac 1 n\dfrac{\sin^2 nt}{\sin^2 t}$. This is a simple use of the geometric series. It is also known that for any $2\pi$-periodic function $f:S^1\to\Bbb R$, the means of the Fourier partial  sums of $f$ are given by $$F_n\star f(s)=\frac{1}{2\pi}\int_0^{2\pi} F_n(t)f(s-t)dt$$
This is verified by using the definition of the $n$-th Fourier coefficient.
If we let $f$ be the function constantly equal to $1$, we get the partial sums all equal to $1$, so $$1=\frac{1}{2\pi}\int_0^{2\pi}F_n(t)dt$$
This is essentially your result. 
A: Here I prove $\displaystyle\int_0^{2\pi}\frac{\sin^2(2n+1)x}{\sin^2x}dx=2\pi(2n+1).$ 
Since $\displaystyle\sum_{j=1}^n \cos 2jx = -\frac{1}{2}+\frac{\sin(2n+1)x}{2\sin(x)}$, we have that $\displaystyle1+\sum_{j=1}^n \cos 2jx = \frac{\sin(2n+1)x}{\sin(x)}$.
Now take both sides to the power of $2$ and integrate term-by-term. $$\begin{align}I&=\int_0^{2\pi}\left(1+2\sum_{j=1}^n \cos 2jx\right)^2dx\\&=2\pi+4\sum_{j=1}^n \int_0^{2\pi}\cos^2 2jx\,dx+4\sum\sum_{j\neq k} \int_0^{2\pi}\cos 2jx\cos 2kx\,dx+4\sum_{j=1}^n \int_0^{2\pi}\cos 2jx\,dx\end{align}$$
Therefore $\displaystyle\int_0^{2\pi}\frac{\sin^2(2n+1)x}{\sin^2x}dx=2\pi(2n+1).$ 
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\,{\rm I}\pars{n}\equiv\int_{0}^{\pi}{\sin^{2}\pars{nx} \over \sin^{2}\pars{x}}\,\dd x = \verts{n}\pi:\ {\large ?}\,,\qquad
n \in {\mathbb Z}}$.

\begin{align}
&\bbox[10px,#ffe]{\ds{\int_{0}^{\pi}{\sin^{2}\pars{nx} \over \sin^{2}\pars{x}}\,\dd x}}
=\int_{0}^{\pi}{1 - \cos\pars{2nx} \over 1 - \cos\pars{2x}}\,\dd x
=\half\int_{0}^{2\pi}{1 - \cos\pars{nx} \over 1 - \cos\pars{x}}\,\dd x
\\[5mm]&=\half\,\Re\int_{0}^{2\pi}
{1 + \ic\verts{n} x- \expo{\ic\verts{n}x} \over 1 - \cos\pars{x}}\,\dd x
=\half\,\Re\oint_{\verts{z}\ =\ 1}
{1 + \verts{n}\ln\pars{z} - z^{\verts{n}} \over 1 - \pars{z^{2} + 1}/\pars{2z}}
\,{\dd z \over \ic z}
\\[5mm]&=-\,\Im
\oint_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}0\ <\ \,{\rm Arg}\pars{z}\ <\ \pi}}
{1 + \verts{n}\ln\pars{z} - z^{\verts{n}} \over \pars{1 - z}^{2}}\,\dd z
\\[5mm]&=\Im\braces{\!\!%
\int_{-1}^{0}\!{1 + \verts{n}\bracks{\ln\pars{-x} + \ic\pi} - x^{\verts{n}}
\over \pars{x - 1}^{2}}\,\dd x
+\int_{0}^{-1}\!{1 + \verts{n}\bracks{\ln\pars{-x} - \ic\pi} - x^{\verts{n}}
\over \pars{x - 1}^{2}}\,\dd x\!\!}
\\[5mm]&=\Im\bracks{2\verts{n}\pi\ic\int_{-1}^{0}{\dd x \over \pars{x - 1}^{2}}}
=2\verts{n}\pi\bracks{-\,{1 \over x - 1}}_{-1}^{0}
=2\verts{n}\pi\pars{1 - \half} =\
\bbox[10px,#ffe,border:1px solid #000]{\ds{\verts{n}\pi}}
\end{align}
A: Untilize the factorization
$a^n-a^{-n}=( a-a^{-1})(a^{n-1} + a^{n-3} + \dots + a^{-n+1} )$
$$ I_n=\int_0^\pi \frac{\sin^2 nx}{\sin^2 x} dx
 =\int_0^\pi \left(\frac{e^{inx}-e^{-inx}}{e^{ix}-e^{-ix}} \right)^2dx
=\int_0^\pi \bigg(\sum_{1\le k\le n} e^{i(n+1-2j)x}\bigg)^2dx\\
$$
and recognize that $\int_0^\pi e^{2i m x}dx =0$, except for $m=0$
\begin{align}
&I_n = \int_0^\pi \sum_{1\le k,j\le n}^{k+j=n+1}e^{i2(n+1-k-j)x}dx 
=\int_0^\pi n\ dx=n\pi
\end{align}
A: As $\sin^2(n+1)x-\sin^2nx=\sin x\sin(2n+1)x,$
$$J(n)=I(n+1)-I(n)=\int_0^\pi\frac{\sin(2n+1)x}{\sin x}dx$$
As $\sin(2m+1)x-\sin(2m-1)x=2\sin x\cos2m x,$
Again, $$J(m)-J(m-1)=2\int_0^\pi\cos2mx\ dx$$
Now for $m\ne0,$ $$\int_0^\pi\cos2mx\ dx=\frac{\sin2mx}{2m}\mid_0^\pi=0$$
$$\implies J(m)=J(m-1)=\cdots=J(0)=\int_0^\pi\frac{\sin(2\cdot0+1)x}{\sin x}dx=(\pi-0)$$
$\implies I(n+1)-I(n)=\pi$ for $n>0$
and $I(1)=\cdots=\pi$
A: Considering the difference
$$
\begin{aligned}
I_{n}-I_{n-1}&=\int_{0}^{\pi} \frac{\sin ^{2}(n x)-\sin ^{2}(n-1) x}{\sin ^{2} x} d x \\
&=\frac{1}{2} \int_{0}^{\pi} \frac{(1-\cos 2 n x)-(1-\cos 2(n-1) x)}{\sin ^{2} x} d x \\
&=\frac{1}{2} \int_{0}^{\pi} \frac{\cos 2(n-1) x-\cos 2 nx}{\sin ^{2} x} d x \\
&=\int_{0}^{\pi} \frac{\sin x \sin (2 n-1) x}{\sin ^{2} x} d x \\
&=\int_{0}^{\pi} \frac{\sin (2 n-1) x}{\sin x} d x \\
&=\pi,
\end{aligned}
$$
where $\pi$ in the last line comes from my post.
Now we can conclude that $$
\begin{aligned}
I_{n}&=\left(I_{n}-I_{n-1}\right)+\left(I_{n-1}-I_{n-2}\right)+\cdots+\left(I_{2}-I_{1}\right) +I_{1} \\
&=(n-1) \pi+\int_{0}^{\pi} 1 d x\\&=n\pi 
\end{aligned}
$$
